# Semi-formal language - Universe has at least three elements

First of all I would like to construct a semi formal sentence, such that the universum has at least three elements. My attempt:

$$\exists x\exists y\exists z (x\not=y\wedge y\not=z\wedge x\not=z)$$

Secondly, is there a (possibly infinite) set of sentences $T$ which has the infinite structures as models? I think it has something to do with Tarskis definition of truth. I use the following notation: $M\vDash \phi$ means $M$ satifies $\phi$, i.e the sentence $\phi$ is valid in $M$

Thirdly, how would you argue that there is no sentence $\phi$ which has the finite structures as models, I mean without a concret proof?

• What do you mean, "without a concrete proof"? Apr 24, 2013 at 19:21
• Can you write a sentence saying that the universe has at least 83 objects? Do you see what this has to do with the second question? Apr 24, 2013 at 19:22
• Well I would do it in the same way as in the first part, instead of taking $x,y,z$ I take $x_1,...,x_{83}$. I have problems seeing the relation to the second part, may you could help me. Without a concret proof I mean a simple argument which makes it somehow clear that there cant be no such sentence, but still not a correct mathematically proof Apr 24, 2013 at 19:25

3. The compactness theorem says that if $\Sigma$ is an inconsistent set of sentences, then there's a finite subset of $\Sigma$ which is inconsistent. This implies [the proof is given in the Wikipedia entry] that if $\varphi$ has arbitrary large finite models, it has an infinite model. So there can't be a $\varphi$ which has all finite structures (however large) as models but no infinite models. So what you need to understand here is compactness ...
• If you want to avoid the compactness theorem for some reason, you could take an non-principal ultraproduct of a sequence of increasingly large models for $\Sigma$ and check that the resulting structure is infinite and is a model of $\Sigma$. ...but this requires knowing about ultraproducts, and moreover it's cheating in a way, since this is one way to prove the compactness theorem. Apr 25, 2013 at 16:41